Python def vowelCount() creating a dictionary

Question

I have to define a function vowelCount(). The input is a list of words and I have to return a dictionary which returns 3 keys. They are 'more consonants' which contains words which have more consonants than vowels, 'more vowels' which have more vowels and 'half vowels' which have equal amount of both.

Here is my code so far:

def voewlCount(wordList):
    myDict = {}
    vowelList = 'AEIOUaeiou'
    contents = wordList.split()
    for word in wordsList:
        if vowelList in wordList == word:
           myDict.append('half vowels')
        elif vowelList in wordList > word:
        myDict.append('more vowels')
    else:
        myDict.append('mostly consasants')

I am getting error messages when I run the shell, saying it is an attribute error sating that a dict has no attribute 'append'

I corrected my code but I am still having issues...here is my new code , Thank You for the help

def vowelContent(wordList):
myDict = {'more consonants':[],'more vowels':[],'half vowels':[]}
vowels = 'aeiouAEIOU'
for word in wordList:
    if vowels in wordList < word:
        myDict['more consonants'].append(word)
    elif vowels in wordLists > word:
        myDict['more vowels'].append(word)
    else:
        myDict['half vowels'].append(word)
return myDict

say = ['do', 'you','know','the','definition','of','insanity','or','being','insane'] print(vowelContent(say))

When I print the function, all the words from the list above are put into the 'more consonants' key

Answer 1

Here's some framework to help you get started. You can fill in the logic that I've left out.

def helper(word):
  """returns the number of vowels and consonants in the word, respectively"""
  # you fill this in
  return n_vowels, n_consonants

def voewlCount(wordList):  #sic
  result = {'more consonants': [], 'more vowels': [], 'half vowels': []}
  for word in wordList:
    nv, nc = helper(word)
    if #something:
      result['more consonants'].append(word)
    elif #something_else:
      result['more vowels'].append(word)
    elif #the other thing:
      result['half vowels'].append(word)
    else:
      # well this can never happen (or can it)?
  return result